1(a)
given #tan2theta=-20/21#
Let #tantheta =x #
So using the identity #tan2theta=(2tantheta)/(1-tan^2theta)# we get
#-20/21=(2x)/(1-x^2)#
#=>-10/21=x/(1-x^2)#
#=>10x^2-21x-10=0#
#=>10x^2-25x+4x-10=0#
#=>5x(2x-5)+2(2x-5)=0#
#=>(2x-5)(5x+2)=0#
So #x=5/2 and x=-2/5#
Hence #tantheta=5/2 and tantheta=-2/5#
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1(b)
given #tan2theta=-36/77#
Let #tantheta =x #
So using the identity #tan2theta=(2tantheta)/(1-tan^2theta)# we get
#36/77=(2x)/(1-x^2)#
#=>18/77=x/(1-x^2)#
#=>18x^2+77x-18=0#
#=>18x^2+81x-4x-18=0#
#=>9x(2x+9)-2(2x+9)=0#
#=>(2x+9)(9x-2)=0#
So #x=-9/2 and x=2/9#
Hence #tantheta=-9/2 and tantheta=2/9#
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2(a)
#cos2theta+5costheta=2#
#=>2cos^2theta-1+5costheta-2=0#
#=>2cos^2theta+5costheta-3=0#
#=>2cos^2theta+6costheta-costheta-3=0#
#=>(2costheta-1)(costheta+3)=0#
So #costheta=-3->"not possible"#
when
#(2costheta-1)=0#
#=>costheta=1/2=cos60^@=cos(360-60)^@=cos300^@#
Hence #theta=60^@ and theta=300^@#
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2(b)
#2sec2x-cot2x=tan2x#
#=>2sec2x=tan2x+cot2x#
#=>2sec2x=(sin2x)/(cos2x)+(cos2x)/(sin2x)#
#=>2sec2x=(sin^2 2x+cos^2 2x)/(sin2xcos2x)=csc2xsec2x#
#=>2sec2x-csc2xsec2x=0#
#=>sec2x(2-csc2x)=0#
#sec2x=0-> "not possible"#
when #(2-csc2x)=0#
#sin2x=1/2=sin30^@=sin(180-30)=sin150^@#
Hence #x= 15^@ and x= 75^@#
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2(c)
#2tan(theta/2)+3tantheta=0#
#=>2tan(theta/2)+(6tan(theta/2))/(1-tan^2(theta/2))=0#
#=>2tan(theta/2)(1+3/(1-tan^2(theta/2)))=0#
#=>2tan(theta/2)(4-tan^2(theta/2))=0#
when
#tan(theta/2)=0=tan(o^@)=tan180^@#
Hence #theta = 0^@ and theta = 360^@#
#(4-tan^2(theta/2))=0#
#=>tan(theta/2)=+-2#
when
#=>tan(theta/2)=+2=tan63.43^@#
#=>theta =126.86^@#
when
#=>tan(theta/2)=-2=-tan63.43^@=tan(180-63.43)#
#=>tan(theta/2)=tan116.57^@#
#=>theta =233.14^@#
